# [R] ? extended rep()

Gabor Grothendieck ggrothendieck at gmail.com
Mon Oct 20 23:50:42 CEST 2008

```Here is one other solution:

x <- 0:1
times <- 3:6
rep(x + 0*times, times)

This solution also works if the length of times is not a whole
number of lengths of x but in that case it does give a warning
which seems reasonable since that is the way recycling in
R works elsewhere too.

On Mon, Oct 20, 2008 at 5:17 PM, Gabor Grothendieck
<ggrothendieck at gmail.com> wrote:
> Try this:
>
> with(data.frame(x = 0:1, times = 3:6), rep(x, times))
>
> or even shorter:
>
> do.call(rep, data.frame(x = 0:1, times = 3:6))
>
>
> On Mon, Oct 20, 2008 at 4:57 PM, Ted Harding
> <Ted.Harding at manchester.ac.uk> wrote:
>> Hi Folks,
>> I'm wondering if there's a compact way to achieve the
>> following. The "dream" is that, by analogy with
>>
>>  rep(c(0,1),times=c(3,4))
>> # [1] 0 0 0 1 1 1 1
>>
>> one could write
>>
>>  rep(c(0,1),times=c(3,4,5,6))
>>
>> which would produce
>>
>> # [1] 0 0 0 1 1 1 1 0 0 0 0 0 1 1 1 1 1 1
>>
>> in effect "recycling" x through 'times'.
>>
>> The objective is to produce a vector of alternating runs of
>> 0s and 1s, with the lengths of the runs supplied as a vector.
>> Indeed, more generally, something like
>>
>>  rep(c(0,1,2), times=c(1,2,3,2,3,4))
>> # [1] 0 1 1 2 2 2 0 0 1 1 1 2 2 2 2
>>
>> Suggestions appreciated! With thanks,
>> Ted.
>>
>> --------------------------------------------------------------------
>> E-Mail: (Ted Harding) <Ted.Harding at manchester.ac.uk>
>> Fax-to-email: +44 (0)870 094 0861
>> Date: 20-Oct-08                                       Time: 21:57:15
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